JP's Blog
Advent of Code: Day 9
https://blog.jverkamp.com/2015/12/09/advent-of-code-day-9/
2015-12-09Part 1: Given a list of distances between cities of the form London to Dublin = 464, calculate the shortest route that visits each city exactly once.
routes = collections.defaultdict(
lambda : collections.defaultdict(
lambda : float("inf")
)
)
for line in sys.stdin:
src, dst, dist = re.match(r'(\w+) to (\w+) = (\d+)', line).groups()
dist = int(dist)
routes[src][dst] = dist
routes[dst][src] = dist
best_length, best_ordering = min(
(sum(
routes[src][dst]
for src, dst in zip(ordering, ordering[1:])
), ordering)
for ordering in itertools.permutations(routes.keys())
)
print(best_length)
There are a few neat tricks here that I’ve used. First routes is defined as nested defaultdict, with an eventual default value of float("inf"). This solves two problems:
- We don’t have to explicitly check if a station exists before adding a route to it:
python routes[src][dst] = distrather thanpython if src in routes: routes[src][dst] = dist else: routes[src] = {dst: dist} - Any missing routes will have an infinite distance, which will work correctly with
+andmin.
We add each distance to both routes[src][dst] and routes[dst][src] so that we don’t have to worry about ordering when we calculate full routes. The other way to do this would be to sort src and dst so that src < dst is always true. I think this way is a little cleaner.
Next, we use a bunch of tools to calculate the shortest route.
First, itertools.permutations will give us every possible ordering:
>>> pprint.pprint(list(itertools.permutations(routes.keys())))
[('London', 'Belfast', 'Dublin'),
('London', 'Dublin', 'Belfast'),
('Belfast', 'London', 'Dublin'),
('Belfast', 'Dublin', 'London'),
('Dublin', 'London', 'Belfast'),
('Dublin', 'Belfast', 'London')]
Next, zip over ordering and ordering[1:] will give us the pairs of stations (since zip after exhausting its shortest argument):
>>> ordering = ('Dublin', 'Belfast', 'London')
>>> pprint.pprint(list(zip(ordering, ordering[1:])))
[('Dublin', 'Belfast'), ('Belfast', 'London')]
Next, we can get the distance for each pairing and sum them all up. This is where float("inf") really comes in handy (although in this smaller example, we don’t need it):
>>> orderings = (routes[src][dst] for src, dst in zip(ordering, ordering[1:]))
>>> pprint.pprint(list(orderings))
[141, 518]
>>> pprint.pprint(sum(orderings))
659
That, we wrap up in a tuple of (distance, ordering) so that they are sortable. Then, apply min to that to find the route with the minimum distance and unpack the tuple again.
And that’s it. Minimum distance.
It’s certainly a brute force solution in that it will try every possible route to find the shortest one. There are probably a few dynamic programming tricks that could cut that down a bit. On the other hand, the input is relatively short (28 connections between 8 stations for a total of 40320 orderings), so just do them all.
Part 2: Find the longest route.
Just change min to max.
This wouldn’t work if there wasn’t a connection listed between every possible station (there is in my given input, since 8 stations will have 8*7/2 = 28 connections). That’s solveable though. Just use float("-inf") for the default distance, so that any routes with invalid stations will have an infinitely small distance.
This was a pretty cool problem again!
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标签：src,Code,ordering,Dublin,dst,Advent,routes,London,2015 From： https://www.cnblogs.com/z-cm/p/17085949.html